International Journals of Advanced Research in
Computer Science and Software Engineering
ISSN: 2277-128X (Volume-7, Issue-6)
Research Article
June
2017
Existence of Nonoscillatory Solutions of First-order Neutral
Dynamic Equations
A. George Maria Selvam
Sacred Heart College,
Tirupattur-635601, S. India
M. Paul Loganathan
Dravidian University,
Kuppam, S. India
S. Janci Rani
Sacred Heart College,
Tirupattur-635601, S. India
Abstract— In this paper, we study the existence of nonoscillatory solution of first-order neutral dynamic equations
with delay and advance terms on Time Scales. Some sufficient conditions for the existence of positive solutions are
obtained. We use the Banach contraction principle to prove our results.
Keywords— Dynamic equations, nonoscillation, positive solution, Banach contraction principle.
I. INTRODUCTION
In this paper we consider a first-order neutral dynamic equation
[ x(n)  P1 (n) x(n  1 )  P2 (n) x(n   2 )]  Q1 (n) x(n  1 )  Q2 (n) x(n   2 )  0 (1.1)
Where P1 , P2  C ([t0 , ), R), Q1 , Q2  C ([t0 , ),[0, )),1, 2  0, and  1 ,  2  0.
Let m  max{1 , 1}. We give some new criteria for the existence of non-oscillatory solutions of (1.1). Recently,
the existence of non-oscillatory solutions of neutral differential equations and difference equations have been
investigated by many authors see books [1,2,9,11] and papers [5,6,8,10,12,13] and the references contained therein.
The theory of time scales, which has recently received a lot of attention, was introduced by Stefan Hilger in his
Ph.D. Thesis [7] in order to unify continuous and discrete analysis. A time scale 𝕋 is an arbitrary nonempty closed subset
of the reals, and scale is equal to the reals or to the integers represent the classical theories of differential and of
difference equations. Many other interesting time scales exist, and they give rise to plenty of applications, among them
the study of population dynamic models (see [3]). A book on the subject of time scales by Bohner and Peterson [3,4]
summarizes and organizes much of the time scale calculus. A solution of the dynamic equation (1.1) is called eventually
positive if there exists a positive integer
that x(n)  0 for
n0 such that x(n)  0 for n  N (n0 ). If there exists a positive integer n0 such
n  N (n0 ), then (1.1) is called eventually negative.
The solution of the dynamic equation (1.1) is said to be oscillatory if it is neither eventually positive nor eventually
negative. Otherwise, it is called nonoscillatory.We need the following important theorem to prove out main results.
Theorem 1.1 (Banach’s Contraction Mapping Principle). A contraction mapping on a complete metric space has exactly
one fixed point.
II. MAIN THEOREMS
To show that an operator 𝑆 satisfies the conditions for the contraction mapping principle, we consider different cases
for the ranges of the coefficients 𝑃1 (𝑡) and𝑃2 (𝑡).
Theorem 2.1 Assume that 1  p1  P1 (n)  0,0  P2 (n)  p2  1  p1 and


 Q ( s)s  ,  Q ( s)s  
1
t0
(2.1)
2
t0
Then (1.1) has a bounded non-oscillatory solution.
Proof: Because of (2.1) we can choose
n1  n0 ,
n1  n0  max{1  1} (2.2)
Sufficiently large such that

 Q (s)s 
1
t
(1  p1 ) M 6  
, n  n1 ,
M6
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(2.3)
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ISSN: 2277-128X (Volume-7, Issue-6)

 Q2 (s)s 
  p2 M 6  M 5
M6
t
, n  n1 ,
(2.4)
Where M 5 and M 6 are positive constants such that
M 5  p2 M 6  (1  p1 )M 2 and   (M 5  p2 M 6 ,(1  p1 )M 2 ) .
Let
ln0 be the set of all real sequence with the norm || x || sup | x(n) | . Then ln0 is a Banach space. We define a
closed, bounded and convex subset  of
ln0 as follows
  {x  ln0 : M 5  x(n)  M 6 , n  n0}.
Define a mapping
S :   ln0 as follows
  P1 (n ) x(n   1 )  P2 (n ) x(n   2 )
 

( Sx )(n )     [Q1 ( s ) x( s   1 )  Q2 ( s ) x( s   2 )]s,n  n1,
 t
( Sx )(n1 ), n0  n  n1
Obviously Sx is continuous. For n  n1' and x , from (2.3) and (2.4), respectively, it follows that

( Sx)(n)    P1 (n) x(n   1 )   Q1 (s) x(s   1 )s
t

   p1M 6  M 6  Q1 ( s)s
t
 (1  p1 ) M 6   
   p1M 6  M 6 

M6


 (Sx)(n)  M 6
Furthermore we have

( Sx)(n)    P2 (n) x(n   2 )   Q2 ( s) x( s   2 )s
t

   p2 M 6  M 6  Q2 (s )s
t
   p2 M 6  M 5 
   p2 M 6  M 6 

M6


(Sx)(n)  M 5
Hence
M 5  (Sx)(n)  M 6 for n  n1
Thus we have proved that ( Sx)(n)  for any x .
This mean that S  . To apply the contraction mapping principle, the remaining is to show that S is a contraction
mapping on  . Thus
x1 , x2  and n  n1 ,
| ( Sx1 )( n )  ( Sx2 )( n) |

   P1 ( n ) x1 ( n   1 )  P2 ( n ) x1 ( n   2 )   [Q1 ( s ) x1 ( s   1 )  Q2 ( s ) x1 ( s   2 )]s
t

(  P1 ( n ) x2 ( n   1 )  P2 ( n ) x2 ( n   2 )   [Q1 ( s ) x2 ( s   1 )  Q2 ( s ) x2 ( s   2 )]s )
t
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Selvam et al., International Journals of Advanced Research in Computer Science and Software Engineering
ISSN: 2277-128X (Volume-7, Issue-6)
 P1 (n) | x1 (n   1 )  x2 (n   1 ) |  P2 (n) | x1 (n   2 )  x2 (n   2 ) |


  Q1 ( s) | x1 ( s   1 )  x2 ( s   1 ) | s   Q2 ( s) | x1 ( s   2 )  x2 ( s   2 ) | s
t
t


t
t
  p1 x1  x2  p2 x1  x2   Q1 (s) x1  x2 s   Q2 (s) x1  x2 s


   p1  p2   Q1 ( s)s   Q2 ( s)s  x1  x2
t
t





(1  p1 ) M 6     p2 M 6  M 5 
   p1  p2 

 x1  x2
M6
M6


M  M5
 6
 x1  x2 
M6
 5 x1  x2
Where
5  1 
M5
. This implies that
M6
(Sx1 )(n)  ( Sx2 )(n)  5 x1  x2
Thus we have to proved that S is a contraction mapping on  . In fact
x1 , x2  and n  n1 we have
(Sx1 )(n)  (Sx2 )(n)  p(n) x1 (n  1 )  x2 (n  1 )  5 x1  x2
Since 0  5  1. We conclude that S is a contraction mapping on  . Thus S has a unique fixed point which is a
positive and bounded solution of (1.1). This completes the proof.
Theorem 2.2.Assume that 1  p1  P1 (n)  0, 1  p1  p2  P2 (n)  0 and (2.1) hold, then (1.1) has a bounded
non-oscillatory solution.
Proof: Because of (2.1), we can choose

 Q (s)s 
1
t

 Q (s)s 
2
t
Where
n1  n0 sufficiently large satisfying (2.2) such that
(1  p1  p2 ) N6  
, n  n1 ,
N2
  N5
N6
, n  n1
(2.5)
(2.6)
N 5 and N 6 are positive constants such that
N5  (1  p1  p2 ) N6 and   ( N5 ,(1  p1  p2 ) N6 ) .
Let
ln0 be the set of all real sequence with the norm || x || sup | x(n) | . Then ln0 is a Banach space. We define a
closed, bounded and convex subset  of
ln0 as follows
  {x  ln0 : N5  x(n)  N6 , n  n0 }.
Define a mapping
S :   ln0 as follows
  P1 (n) x(n   1 )  P2 (n) x(n   2 )
 

( Sx)(n)    [Q1 ( s) x( s   1 )  Q2 ( s) x( s   2 )]s,n  n1 ,
 t
( Sx)(n ), n  n  n .
1
0
1

Obviously Sx is continuous. For
n  n1' and x , from (2.5) and (2.6), respectively, it follows that
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Selvam et al., International Journals of Advanced Research in Computer Science and Software Engineering
ISSN: 2277-128X (Volume-7, Issue-6)

( Sx)(n)    P1 (n) x(n   1 )  P2 (n) x(n   2 )   Q1 (s) x( s   1 )s
t

   p1 N6  p2 N 6  N 6  Q1 (s )s
t
 (1  p1  p2 ) N6   
   p1 N6  p2 N 6  N 6 

N6


 (Sx)(n)  N6
Furthermore we have

( Sx)(n)     Q2 ( s) x( s   2 )s
t

   N 6  Q2 ( s)s
t
   N5 
   N6 

 N6 
(Sx)(n)  N5
Hence
N5  (Sx)(n)  N6 for n  n1
Thus we have proved that ( Sx)(n)  for any x .
This mean that S  . To apply the contraction mapping principle, the remaining is to show that S is a contraction
mapping on  . Thus
x1 , x2  and n  n1 ,
| ( Sx1 )( n)  ( Sx2 )( n) |

   P1 (n ) x1 (n   1 )  P2 (n ) x1 ( n   2 )   [Q1 ( s ) x1 ( s   1 )  Q2 ( s) x1 ( s   2 )]s
t

(  P1 ( n ) x2 ( n   1 )  P2 ( n ) x2 ( n   2 )   [Q1 ( s ) x2 ( s   1 )  Q2 ( s) x2 ( s   2 )]s )
t
 P1 (n) | x1 (n   1 )  x2 (n  1 ) |  P2 (n) | x1 (n   2 )  x2 (n   2 ) |


  Q1 ( s) | x1 ( s   1 )  x2 ( s   1 ) | s   Q2 ( s) | x1 ( s   2 )  x2 ( s   2 ) | s
t
t


t
t
  p1 x1  x2  p2 x1  x2   Q1 (s) x1  x2 s   Q2 (s) x1  x2 s




   p1  p2   Q1 ( s)s   Q2 ( s)s  x1  x2
t
t



(1  p1  p2 ) N6     N5 
   p1  p2 

 x1  x2
N6
N6 

N  N5
 6
 x1  x2 
N6
 6 x1  x2
Where
6  1 
N5
. This implies that
N6
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ISSN: 2277-128X (Volume-7, Issue-6)
(Sx1 )(n)  ( Sx2 )(n)  6 x1  x2
Thus we have to proved that S is a contraction mapping on  . In fact
x1 , x2  and n  n1 we have
(Sx1 )(n)  (Sx2 )(n)  p(n) x1 (n  1 )  x2 (n  1 )  6 x1  x2
Since 0  6  1. We conclude that S is a contraction mapping on  . Thus S has a unique fixed point which is a
positive and bounded solution of (1.1). This completes the proof.
Theorem 2.3. Assume that   p10  P1 (n)  p1  1,0  P2 (n)  p2   p1  1 and (2.1) hold, then (1.1) has a
bounded non-oscillatory solution.
Proof: In view of (2.1), we can choose
n1  n0 sufficiently large satisfying
n1  1  n0  1 , (2.7)
such that

 Q (s)s 
p10 M 7  
, n  n1 ,
(2.8)
( p1  1  p2 ) M 8  
, n  n1 ,
M8
(2.9)
1
M8
t

 Q (s)s 
2
t
where M 7 and M 8 are positive constants such that
 p10 M 7  ( p1  1  p2 )M 8 and   ( p10 M 7 ,( p1  1  p2 )M 8 ) .
Let
ln0 be the set of all real sequence with the norm || x || sup | x(n) | . Then ln0 is a Banach space. We define a
closed, bounded and convex subset  of
ln0 as follows
  {x  ln0 : M 7  x(n)  M 8 , n  n0 }.
Define a mapping
S :   ln0 as follows
 1
 P (n   ) {  x(n   1 )  P2 (n   1 ) x(n   1   2 )
1
 1
 
( Sx)(n)    [Q1 ( s) x( s   1 )  Q2 ( s ) x( s   2 )]s,n  n1 ,
 t 1
( Sx)(n ), n  n  n .
1
0
1


Clearly Sx is continuous. For n  n1' and x , from (2.9) and (2.8), respectively, it follows that


1 
( Sx)(n) 
   x(n   1 )  P2 (n   1 ) x(n   1   2 )   Q2 ( s) x( s   2 )s 

P1 (n   1 ) 
t 1





1 


M

p
M

M

8
2
8
8  Q2 ( s )s 
p1 
t

 ( p1  1  p2 ) M 8    
1 
   M 8  p2 M 8  M 8 
 
p1 
M8


 (Sx)(n)  M 8
Furthermore we have
( Sx)(n) 


1 
    Q1 ( s) x( s   1 )s 

P1 (n   1 ) 
t 1

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ISSN: 2277-128X (Volume-7, Issue-6)


1 

   M 8  Q1 ( s)s 
P10 
t


 p10 M 7    
1 
   M 8 
 
p10 
M8


 (Sx)(n)  M 7
Hence
M 7  (Sx)(n)  M 8 for n  n1
Thus we have proved that ( Sx)(n)  for any x .
This mean that S  . To apply the contraction mapping principle, the remaining is to show that S is a contraction
mapping on  . Thus
x1 , x2  and n  n1 ,
| ( Sx1 )(n)  ( Sx2 )(n) |


1
{  x1 (n   1 )  P2 (n   1 ) x1 (n   1   2 )   [Q1 ( s) x1 ( s   1 )  Q2 ( s) x1 ( s   2 )]s}
P1 (n   1 )
t 1



1

{  x2 (n   1 )  P2 (n   1 ) x2 (n   1   2 )   [Q1 ( s ) x2 ( s   1 )  Q2 ( s ) x2 ( s   2 )]s} 

 P1 (n   1 )
t 1


1

| x1 (n  1 )  x2 (n  1 ) |  P2 (n  1 ) | x1 (n  1   2 )  x2 (n  1   2 ) |
P1 (n   1 )



Q1 ( s) | x1 ( s   1 )  x2 ( s   1 ) | s 
t 1

 Q ( s) | x ( s  
2
1
2
)  x2 ( s   2 ) |  s 
t 1



1 
  x1  x2  p2 x1  x2   Q1 ( s) x1  x2 s   Q2 (s) x1  x2 s 
p1 
t
t




1 
 1  p2   Q1 ( s)s   Q2 ( s)s  x1  x2
p1 
t
t

p1 M 7   ( p1  1  p2 ) M 8   
1 
 1  p2  0

 x1  x2
p1 
M8
M8

1  p1 M 7  p1M 8 
  0
 x1  x2
p1 
M8

 7 x1  x2
Where
7  1 
p10 M 7
p1M 8
This implies that
(Sx1 )(n)  ( Sx2 )(n)  7 x1  x2
Thus we have to proved that S is a contraction mapping on  . In fact
x1 , x2  and n  n1 we have
(Sx1 )(n)  (Sx2 )(n)  p(n) x1 (n  1 )  x2 (n  1 )  7 x1  x2
Since 0  7  1. We conclude that S is a contraction mapping on  . Thus S has a unique fixed point which is a
positive and bounded solution of (1.1). This completes the proof.
Theorem 2.4. Assume that   p10  P1 (n)  p1  1, p1  1  p2  P2 (n)  0 and (2.1) hold, then (1.1) has a
bounded non-oscillatory solution.
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Proof: In view of (2.1), we can choose n1  n0 sufficiently large satisfying
n1  1  n0  1, (2.10)
such that

 Q1 (s)s 
p10 N 7  p2 N8  
N8
t

 Q (s)s 
2
t
Where
, n  n1 , (2.11)
( p1  1) N8  
, n  n1 , (2.12)
N8
N 7 and N8 are positive constants such that
 p10 N7  p2 N8  ( p1  1) N8 and   ( p10 N7  p2 N8 ,( p1  1) N8 ) .
Let
ln0 be the set of all real sequence with the norm || x || sup | x(n) | . Then ln0 is a Banach space. We define a
closed, bounded and convex subset  of
ln0 as follows
  {x  ln0 : N7  x(n)  N8 , n  n0 }.
Define a mapping
S :   ln0 as follows
 1
 P (n   ) {  x(n   1 )  P2 (n   1 ) x(n   1   2 )
1
 1


( Sx)(n)    [Q1 ( s) x( s   1 )  Q2 ( s ) x( s   2 )]s,n  n1 ,
 t 1
( Sx)(n ), n  n  n .
1
0
1


Clearly Sx is continuous. For
( Sx)(n) 


n  n1' and x , from (2.8) and (2.9), respectively, it follows that


1 
   x(n   1 )   Q2 ( s) x( s   1 )s 

P1 (n   1 ) 
t 1



1 


N

N

8
8  Q2 ( s )s 
p1 
t

 ( p1  1) N8    
1
   N8  N8 
 
p1 
N8


 (Sx)(n)  N8
Furthermore we have


1 
( Sx)(n) 
   P2 (n  1 ) x(n  1   2 )   Q1 ( s) x( s   1 )s 

P1 (n   1 ) 
t 1




1 


p
N

N

2 8
8  Q1 ( s )s 
p10 
t


 p10 N 7  p2 N8    
1 
   p2 N8  N8 
 
p10 
N8


 (Sx)(n)  N7
Hence
N7  (Sx)(n)  N8 for n  n1
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Selvam et al., International Journals of Advanced Research in Computer Science and Software Engineering
ISSN: 2277-128X (Volume-7, Issue-6)
Thus we have proved that ( Sx)(n)  for any x .
This mean that S  . To apply the contraction mapping principle, the remaining is to show that S is a contraction
mapping on  . Thus
x1 , x2  and n  n1 ,
| ( Sx1 )(n)  ( Sx2 )(n) |

1

{  x1 (n   1 )  P2 (n   1 ) x1 (n   1   2 )   [Q1 ( s) x1 ( s   1 )  Q2 ( s) x1 ( s   2 )]s}
P1 (n   1 )
t 1



1

{  x2 (n   1 )  P2 (n   1 ) x2 (n   1   2 )   [Q1 ( s ) x2 ( s   1 )  Q2 ( s ) x2 ( s   2 )]s} 

 P1 (n   1 )
t 1


1

| x1 (n  1 )  x2 (n  1 ) |  P2 (n  1 ) | x1 (n  1   2 )  x2 (n  1   2 ) |
P1 (n   1 )



Q1 ( s) | x1 ( s   1 )  x2 ( s   1 ) | s 
t 1

 Q ( s) | x ( s  
2
1
2
)  x2 ( s   2 ) |  s 
t 1



1 
 x1  x2  p2 x1  x2   Q1 ( s) x1  x2 s   Q2 ( s) x1  x2 s 
p1 
t
t




1 
 1  p2   Q1 ( s)s   Q2 ( s)s  x1  x2
p1 
t
t



p10 N7  p2 N8   ( p1  1) N8   
1 

1  p2 
 x1  x2
p1 
N8
N8


1  p10 N 7  p1 N8 

 x1  x2
p1 
N8

 8 x1  x2
Where 8
 1
p10 N 7
p1 N8
. This implies that
(Sx1 )(n)  ( Sx2 )(n)  8 x1  x2
Thus we have to proved that S is a contraction mapping on  . In fact
x1 , x2  and n  n1 we have
(Sx1 )(n)  (Sx2 )(n)  p(n) x1 (n  1 )  x2 (n  1 )  8 x1  x2
Since 0  8  1. We conclude that S is a contraction mapping on  . Thus S has a unique fixed point which is a
positive and bounded solution of (1.1). This completes the proof.
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